Skip to main content
Back to Differentiation
JEE Main 2019
Differentiation
Differentiation
Hard

Question

If y(θ)=2cosθ+cos2θcos3θ+4cos2θ+5cosθ+2y(\theta)=\frac{2 \cos \theta+\cos 2 \theta}{\cos 3 \theta+4 \cos 2 \theta+5 \cos \theta+2}, then at θ=π2,y+y+y\theta=\frac{\pi}{2}, y^{\prime \prime}+y^{\prime}+y is equal to :

Options

Solution

y(θ)=2cosθ+cos2θcos3θ+4cos2θ+5cosθ+2=2cos2θ+2cosθ14cos3θ+8cos2θ+2cosθ2=2cos2θ+2cosθ1(2cos2θ+2cosθ1)(2cosθ+2)=12(1+cosθ)=14cos2θ/2=sec2θ/24y(θ)=14(2secθ2secθ2tanθ212)=14sec2θ2tanθ2\begin{aligned} & y(\theta)=\frac{2 \cos \theta+\cos 2 \theta}{\cos 3 \theta+4 \cos 2 \theta+5 \cos \theta+2} \\ & =\frac{2 \cos ^2 \theta+2 \cos \theta-1}{4 \cos ^3 \theta+8 \cos ^2 \theta+2 \cos \theta-2} \\ & =\frac{2 \cos ^2 \theta+2 \cos \theta-1}{\left(2 \cos ^2 \theta+2 \cos \theta-1\right)(2 \cos \theta+2)} \\ & =\frac{1}{2(1+\cos \theta)}=\frac{1}{4 \cos ^2 \theta / 2}=\frac{\sec ^2 \theta / 2}{4} \\ & y^{\prime}(\theta)=\frac{1}{4}\left(2 \sec \frac{\theta}{2} \cdot \sec \frac{\theta}{2} \cdot \tan \frac{\theta}{2} \cdot \frac{1}{2}\right) \\ & =\frac{1}{4} \sec ^2 \frac{\theta}{2} \cdot \tan \frac{\theta}{2} \end{aligned} y(θ)=14(tanθ2)(sec2θ2tanθ2)+14sec2θ2sec2θ212y^{\prime \prime}(\theta)=\frac{1}{4}\left(\tan \frac{\theta}{2}\right)\left(\sec ^2 \frac{\theta}{2} \cdot \tan \frac{\theta}{2}\right) +\frac{1}{4} \sec ^2 \frac{\theta}{2} \cdot \sec ^2 \frac{\theta}{2} \cdot \frac{1}{2}  at θ=π2,y(θ)=12,y(θ)=12,y(θ)=1y+y+y=2\begin{aligned} & \text { at } \theta=\frac{\pi}{2}, y(\theta)=\frac{1}{2}, y^{\prime}(\theta)=\frac{1}{2}, y^{\prime \prime}(\theta)=1 \\ & \therefore \quad y+y^{\prime}+y^{\prime \prime}=2 \end{aligned}

Practice More Differentiation Questions

View All Questions